Question

A body at rest falls through a height 'h' with velocity 'V'. If it has to fall down further for its velocity to become three times, the distance travelled in that interval is

• A. $8h$
• B. $6h$
• C. $4h$
• D. $12h$

Hint:

Since distance scales with the square of velocity ($H \propto v^2$), tripling the velocity ($1 \to 3$) requires multiplying the total fall distance from rest by $3^2 = 9$. Subtracting the initial distance $1h$ gives an additional distance of $9h - 1h = 8h$ instantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
An object dropped from rest falls down a vertical distance $h$ and reaches a final velocity $V$. We need to calculate the

additional distance it must fall for its velocity to increase to $3V$.

Step 2: Key Formula or Approach:
We can solve this problem using the third equation of motion for uniform acceleration under gravity ($u=0$): $$v^2 = u^2 + 2gH \implies v^2 = 2gH$$ This shows that the square of the final velocity is directly proportional to the total vertical distance fallen from the release point ($v^2 \propto H$).

Step 3: Detailed Explanation:
Let's establish the equations for the two distinct velocity check points: 1. For the first interval, falling a distance $h$ to reach velocity $V$: $$V^2 = 2gh \implies h = \frac{V^2}{2g}$$ 2. Let $h'$ be the total vertical distance fallen from rest to reach a velocity of $3V$: $$(3V)^2 = 2gh'$$ $$9V^2 = 2gh' \implies h' = \frac{9V^2}{2g}$$ Substitute our expression for $h$ into the equation for $h'$: $$h' = 9 \left(\frac{V^2}{2g}\right) = 9h$$ The question specifically asks for the distance traveled

in that interval (the further distance fallen after reaching velocity $V$): $$\Delta h = h' - h = 9h - h = 8h$$

Step 4: Final Answer:
The additional distance traveled in that interval is $8h$, which corresponds to option (A).